A) \[\frac{m}{M}L\,\cos \theta \]
B) \[\frac{m}{M+m}L\]
C) \[\frac{M+m}{mL\,\cos \theta }\]
D) \[\frac{mL\,\cos \theta }{m+M}\]
Correct Answer: D
Solution :
Here, the x-coordinate of centre of mass of the system remains unchanged when the mass m moved a distance \[L\,\cos \theta ,\] Let the mass \[(m+M)\]moves a distance X in the backward direction. \[\therefore \] \[(m+m)x-m\,L\,\cos \theta =0\] \[X-\frac{mL\cos \theta }{m+M}\] Here, note that the Y-coordinate of the centre of mass would be change.You need to login to perform this action.
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